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\title{\heiti\zihao{2} 习题16.3}
\author{中书君}
\date{\today}
\begin{document}
\maketitle
\section{设$f(x,y,z)$为连续函数,求$\lim\limits_{r\rightarrow 0^+}\dfrac{3}{4\pi r^3}\iiint\limits_{x^2+y^2+z^2\leqslant r^2}f(x,y,z)\mathrm{d}x\mathrm{d}y\mathrm{d}z$}
\textbf{解}\quad
$$
\begin{aligned}
    \dfrac{3}{4\pi r^3}\iiint\limits_{x^2+y^2+z^2\leqslant r^2}f(x,y,z)\mathrm{d}x\mathrm{d}y\mathrm{d}z&=\dfrac{3}{4\pi r^3}f(x_0,y_0,z_0)\int_0^{\pi}\mathrm{d}\varphi\int_0^{2\pi}\mathrm{d}\theta\int_0^1r^2\cos\varphi \mathrm{d}r\\
    &=f(x_0,y_0,z_0)
\end{aligned}
$$
从而由积分中值定理:
$$
\lim\limits_{r\rightarrow 0^+}\dfrac{3}{4\pi r^3}\iiint\limits_{x^2+y^2+z^2\leqslant r^2}f(x,y,z)\mathrm{d}x\mathrm{d}y\mathrm{d}z=\lim\limits_{r\rightarrow 0^+}f(x,y,z)=f(0,0,0)
$$

\section{证明:若函数 $f(x, y, z)$ 在区域 $V$ 中连续且对于任意子区域 $\Omega \subset V$, 都有 $\iiint\limits_{\Omega} f(x, y, z) \mathrm{d} x \mathrm{~d} y \mathrm{~d} z=0$, 则 $f(x, y, z) \equiv 0,(x, y, z) \in V$.}
\textbf{解}\quad
由积分中值定理:
$$
\iiint\limits_{\Omega} f(x, y, z) \mathrm{d} x \mathrm{~d} y \mathrm{~d} z=f(x_0, y_0, z_0)\iiint\limits_{\Omega}  \mathrm{d} x \mathrm{~d} y \mathrm{~d} z=0
$$
从而$f(x_0,y_0,z_0)=0$对于$\Omega$中每一个点都成立(因为是区域).从而$f(x, y, z) \equiv 0,(x, y, z) \in V$.
\section{计算下列积分:}
\subsection{$\iiint\limits_{V}(x y+2 z) \mathrm{d} x \mathrm{~d} y \mathrm{~d} z$, 其中 $V=[2,3] \times[-1,1] \times[0,2]$}
\textbf{解}\quad
在此区域上,$xy$关于$y$是奇函数,且区域关于$xoz$平面对称,从而该项积分为$0$.所以
$$
\begin{aligned}
    \iiint\limits_{V}(x y+2 z) \mathrm{d} x \mathrm{~d} y \mathrm{~d} z&=\iiint\limits_V2z\mathrm{d}x\mathrm{d}y\mathrm{d}z\\
    &=\int_2^3\mathrm{d}x\int_{-1}^1\mathrm{d}y\int_0^22z\mathrm{d}z\\
    &=8
\end{aligned}
$$

\subsection{$\iiint\limits_{V} y z \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z, V$ 是单位球 $x^{2}+y^{2}+z^{2} \leqslant 1$ 在第一卦限内的部分}
\textbf{解}\quad
$$
\begin{aligned}
    \iiint\limits_{V} y z \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z&=\int_0^{\pi/2}\mathrm{d}\varphi\int_0^{\pi/2}\mathrm{d}\theta\int_0^1r^4\sin^2\varphi\cos\varphi\sin\theta\mathrm{d}r\\
    &=\dfrac{1}{5}\cdot\int_0^{\pi/2}\sin^2\varphi\mathrm{d}\sin\varphi\\
    &=\dfrac{1}{15}
\end{aligned}
$$

\subsection{$\iiint\limits_{V} \dfrac{\mathrm{d} x \mathrm{~d} y \mathrm{~d} z}{(1+x+y+z)^{3}}, V$ 是由平面 $x+y+z=1$ 与三个坐标平面所围成的区域}
\textbf{解}\quad
$$
\begin{aligned}
    \iiint\limits_{V} \dfrac{\mathrm{d} x \mathrm{~d} y \mathrm{~d} z}{(1+x+y+z)^{3}}&=\iint\limits_{\Omega_{xy}}\mathrm{d}x\mathrm{d}y\int_0^{1-x-y}\dfrac{\mathrm{d}z}{(1+x+y+z)^3}\\
    &=-\dfrac{1}{2}\int_0^1\mathrm{d}x\int_0^{1-x}\left[\dfrac{1}{4}-\dfrac{1}{(1+x+y)^2}\right]\mathrm{d}y\\
    &=-\dfrac{1}{2}\int_0^1\dfrac{3}{4}-\dfrac{x}{4}-\dfrac{1}{x+1}\mathrm{d}x\\
    &=\dfrac{1}{2}\ln 2-\dfrac{5}{16}
\end{aligned}
$$

\section{计算下列积分:}
\subsection{$\iiint\limits_Vz^{2} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z$, 其中 $V$ 是由曲面 $x^{2}+y^{2}+z^{2}=r^{2}$ 和 $x^{2}+y^{2}+z^{2}=2 r z$ 所围成的区域}
\textbf{解}\quad 
\begin{figure}[h]
    \centering
    \includegraphics[scale = 0.3]{include_picture/16.3.1.png}
    \caption{}
\end{figure}
柱坐标变换:$\left\{\begin{array}{l}
    x=\rho\cos\varphi\\
    y=\rho\sin\varphi\\
    z=z
\end{array}\right.$.
$$
\begin{aligned}
    \iiint\limits_Vz^{2} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z&=\int_0^{r/2}\mathrm{d}z\int_0^{2\pi}\mathrm{d}\varphi\int_0^{\sqrt{2rz-z^2}}z^2\rho\mathrm{d}\rho\\
    &+\int_{r/2}^{r}\mathrm{d}z\int_0^{2\pi}\mathrm{d}\varphi\int_0^{\sqrt{r^2-z^2}}z^2\rho\mathrm{d}\rho\\
    &=\pi\int_0^{r/2}z^2(2rz-z^2)\mathrm{d}z+\pi\int_{r/2}^{r}z^2(r^2-z^2)\mathrm{d}z\\
    &=\dfrac{59\pi r^2}{480}
\end{aligned}
$$


\subsection{$\iiint\limits_{V}\left(x^{2}+y^{2}\right)^{2} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z$,其中 $V$ 是由曲面 $z=x^{2}+y^{2}, z=1, z=2$ 所围成的区域}
\textbf{解}\quad
柱坐标变换:$\left\{\begin{array}{l}
    x=\rho\cos\varphi\\
    y=\rho\sin\varphi\\
    z=z
\end{array}\right.$.
$$
\begin{aligned}
    \iiint\limits_{V}\left(x^{2}+y^{2}\right)^{2} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z&=\int_1^2\mathrm{d}z\int_0^{2\pi}\mathrm{d}\varphi\int_0^{\sqrt{z}}\rho^5\mathrm{d}\rho\\
    &=\pi\int_1^2\dfrac{z^3}{3}\mathrm{d}z\\
    &=\dfrac{5\pi}{4}
\end{aligned}
$$

\subsection{$\iiint\limits_{V}\left(x^{2}+y^{2}\right)^{\dfrac{3}{2}} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z$, 其中 $V$ 是曲面 $x^{2}+y^{2}=9, x^{2}+y^{2}=16, z^{2}=x^{2}+y^{2}, z \geqslant 0$}
\textbf{解}\quad
柱坐标变换:$\left\{\begin{array}{l}
    x=\rho\cos\varphi\\
    y=\rho\sin\varphi\\
    z=z
\end{array}\right.$.
截面:
\begin{tikzpicture}
    \draw[->](-0.2,0)--(5,0)node[right]{$x$};
	\draw[->](0,-0.2)--(0,5)node[above]{$y$};
    \draw(3,0)--(3,3);
    \draw(3,3)--(4,4);
    \draw(4,4)--(4,0);
\end{tikzpicture}
$$
\begin{aligned}
    \iiint\limits_{V}\left(x^{2}+y^{2}\right)^{\dfrac{3}{2}} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z&=\iint\limits_{D}\mathrm{d}\varphi\mathrm{d}\rho\int_0^\rho\rho^4\mathrm{d}z\\
    &=\int_0^{2\pi}\mathrm{d}\varphi\int_3^4\mathrm{d}\rho\int_0^\rho\rho^4\mathrm{d}\rho\\
    &=2\pi\int_3^4\rho^5\mathrm{d}\rho\\
    &=\dfrac{3367\pi}{3}
\end{aligned}
$$

\subsection{$\iiint\limits_{V} \sqrt{x^{2}+y^{2}+z^{2}} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z$, 其中 $V=\left\{(x, y, z) \mid x^{2}+y^{2}+z^{2} \leqslant z\right\}$}
\textbf{解}\quad
球坐标变换:$\left\{\begin{array}{l}
    x=\rho\sin\varphi\cos\theta\\
    y=\rho\sin\varphi\sin\theta\\
    z=\rho\cos\varphi
\end{array}\right.$.
由于其区域为圆心在$(0,0,1/2)$,半径为$\dfrac{1}{2}$的球,从而可知$\varphi\in\left[0,\dfrac{\pi}{2}\right]$.
$$
\begin{aligned}
    \iiint\limits_{V} \sqrt{x^{2}+y^{2}+z^{2}} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z&=\int_0^{2\pi}\mathrm{d}\theta\int_0^{\pi/2}\mathrm{d}\varphi\int_0^{\cos\varphi}\rho^3\sin\varphi\mathrm{d}\rho\\
    &=-2\pi\int_0^{\pi/2}\dfrac{\cos^4\varphi}{4}\mathrm{d}\cos\varphi\\
    &=\dfrac{\pi}{10}
\end{aligned}
$$

\subsection{$\iiint\limits_{V} z \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z$, 其中 $V=\left\{(x, y, z) \mid \dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}+\dfrac{z^{2}}{c^{2}} \leqslant 1, z \geqslant 0\right\}$}
\textbf{解}\quad
广义球坐标变换:$\left\{\begin{array}{l}
    x=a\rho\sin\varphi\cos\theta\\
    y=b\rho\sin\varphi\sin\theta\\
    z=c\rho\cos\varphi
\end{array}\right.$,则$\dfrac{\partial(x,y,z)}{\partial(u,v,w)}=abc\rho^2\sin\varphi$.
$$
\begin{aligned}
    \iiint\limits_{V} z \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z&=\int_0^{2\pi}\mathrm{d}\theta\int_0^{\pi/2}\mathrm{d}\varphi\int_0^1abc^2\rho^3\cos\varphi\sin\varphi\mathrm{d}\rho\\
    &=\dfrac{abc^2\pi}{4}
\end{aligned}
$$

\subsection{$\iiint\limits_{V} \sqrt{1-\left(\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}+\dfrac{z^{2}}{c^{2}}\right)} \mathrm{d} x \mathrm{~d} y \mathrm{~d} z$, 其中 $V=\left\{(x, y, z) \mid \dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}+\dfrac{z^{2}}{c^{2}} \leqslant 1\right\}$}
\textbf{解}\quad
广义球坐标变换:$\left\{\begin{array}{l}
    x=a\rho\sin\varphi\cos\theta\\
    y=b\rho\sin\varphi\sin\theta\\
    z=c\rho\cos\varphi
\end{array}\right.$,则$\dfrac{\partial(x,y,z)}{\partial(u,v,w)}=abc\rho^2\sin\varphi$.
$$
\begin{aligned}
    \iiint\limits_{V} \sqrt{1-\left(\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}+\dfrac{z^{2}}{c^{2}}\right)} \mathrm{d} x \mathrm{~d} y \mathrm{~d} z&=abc\int_0^{2\pi}\mathrm{d}\theta\int_0^{\pi}\mathrm{d}\varphi\int_0^1\rho^2\sqrt{1-\rho^2}\mathrm{d}\rho\\
    &=\dfrac{abc\pi^2}{4}
\end{aligned}
$$

\subsection{$\iiint\limits_{V} e^{\sqrt{\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}+\dfrac{z^{2}}{c^{2}}}} \mathrm{d} x \mathrm{d} y \mathrm{d} z$, 其中 $V=\left\{(x, y, z) \mid \dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}+\dfrac{z^{2}}{c^{2}} \leqslant 1\right\}$}
\textbf{解}\quad
广义球坐标变换:$\left\{\begin{array}{l}
    x=a\rho\sin\varphi\cos\theta\\
    y=b\rho\sin\varphi\sin\theta\\
    z=c\rho\cos\varphi
\end{array}\right.$,则$\dfrac{\partial(x,y,z)}{\partial(u,v,w)}=abc\rho^2\sin\varphi$.
$$
\begin{aligned}
    \iiint\limits_{V} e^{\sqrt{\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}+\dfrac{z^{2}}{c^{2}}}} \mathrm{d} x \mathrm{d} y \mathrm{d} z&=abc\int_0^{2\pi}\mathrm{d}\theta\int_0^{\pi}\mathrm{d}\varphi\int_0^1\rho^2\mathrm{e}^{\rho}\sin\varphi\mathrm{d}\rho\\
    &=4\pi abc\int_0^1\rho^2\mathrm{e}^{\rho}\mathrm{d}\rho\\
    &=4\pi abc(\mathrm{e}-2)
\end{aligned}
$$
\subsection{$\int_{0}^{1} \mathrm{~d} x \int_{0}^{\sqrt{1-x^{2}}} \mathrm{~d} y \int_{\sqrt{x^{2}+y^{2}}}^{\sqrt{2-x^{2}-y^{2}}} z^{2} \mathrm{~d} z$}
\textbf{解}\quad
柱坐标变换:
$$
\begin{aligned}
    \int_{0}^{1} \mathrm{~d} x \int_{0}^{\sqrt{1-x^{2}}} \mathrm{~d} y \int_{\sqrt{x^{2}+y^{2}}}^{\sqrt{2-x^{2}-y^{2}}} z^{2} \mathrm{~d} z&=\int_0^{\pi/2}\mathrm{d}\theta\int_0^1\mathrm{d}\rho\int_\rho^{\sqrt{2-\rho^2}}\rho z^2\mathrm{d}z\\
    &=\dfrac{\pi}{15}(2\sqrt{2}-1)
\end{aligned}
$$

\section{利用适当的变量代换,计算积分 $\iiint\limits_{V} x^{2} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z$,其中 $V$ 是由曲面 $z=a y^{2}, z=b y^{2}, y>0$, $z=\alpha x, z=\beta x, z=h(0<a<b,0<\alpha<\beta, h>0)$ 所围成的区域}
\textbf{解}\quad
$$
\begin{aligned}
    \iiint\limits_{V} x^{2} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z&=\int_0^h\mathrm{d}z\int_{z/\beta}^{z/\alpha}\mathrm{d}x\int_{\sqrt{z/b}}^{\sqrt{z/a}}x^2\mathrm{d}y\\
    &=\dfrac{2}{27}\dfrac{h^{9/2}(\beta^3\sqrt{b}-\beta^3\sqrt{a}-\alpha^3\sqrt{b}+\alpha^3\sqrt{a})}{\alpha^3\beta^3\sqrt{ab}}
\end{aligned}
$$

\section{计算由下列曲面围成的立体的体积:}
\subsection{$\left(\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}+\dfrac{z^{2}}{c^{2}}\right)^{2}=\dfrac{x}{h}$}
\textbf{解}\quad
广义球坐标变换:$\left\{\begin{array}{l}
    x=a\rho\sin\varphi\cos\theta\\
    y=b\rho\sin\varphi\sin\theta\\
    z=c\rho\cos\varphi
\end{array}\right.$,则$\dfrac{\partial(x,y,z)}{\partial(u,v,w)}=abc\rho^2\sin\varphi$.

有$\rho^4\leqslant\dfrac{a\rho\sin\varphi\cos\theta}{h}$即$\rho^3\leqslant\dfrac{a\sin\varphi\cos\theta}{h}$.$\sin\varphi\geqslant 0$,从而$\sin\theta\geqslant 0$.
$$
\begin{aligned}
    \iiint\limits_{\Omega}\mathrm{d}x\mathrm{d}y\mathrm{d}z&=abc\int_0^{\pi}\mathrm{d}\varphi\int_{-\pi/2}^{\pi/2}\mathrm{d}\theta\int_0^{\sqrt[3]{\dfrac{a}{h}\sin\varphi\cos\theta}}\rho^2\sin\varphi\mathrm{d}\rho\\
    &=\dfrac{a^2bc}{3h}\int_0^\pi\sin^2\varphi\mathrm{d}\varphi\int_{-\pi/2}^{\pi/2}\cos\theta\mathrm{d}\theta\\
    &=\dfrac{a^2bc\pi}{3h}
\end{aligned}
$$

\subsection{$\left(\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}+\dfrac{z^{2}}{c^{2}}\right)^{2}=\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}$}
\textbf{解}\quad
广义球坐标变换:$\left\{\begin{array}{l}
    x=a\rho\sin\varphi\cos\theta\\
    y=b\rho\sin\varphi\sin\theta\\
    z=c\rho\cos\varphi
\end{array}\right.$,则$\dfrac{\partial(x,y,z)}{\partial(u,v,w)}=abc\rho^2\sin\varphi$.

有$\rho^4\leqslant\rho^2\sin^2\varphi$,即$\rho^2\leqslant\sin^2\varphi$.
$$
\begin{aligned}
    \iiint\limits_{\Omega}\mathrm{d}x\mathrm{d}y\mathrm{d}z&=abc\int_0^{2\pi}\mathrm{d}\theta\int_0^\pi\mathrm{d}\varphi\int_0^{\sin\varphi}\rho^2\sin\varphi\mathrm{d}\rho\\
    &=\dfrac{4\pi abc}{3}\int_0^{\pi/2}\sin^4\theta\mathrm{d}\theta\\
    &=\dfrac{\pi^2 abc}{4}
\end{aligned}
$$





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